Answer:
11.58 L of N₂
Explanation:
We'll begin by calculating the number of mole in 37.2 g of magnesium. This can be obtained as follow:
Mass of Mg = 37.2 g
Molar mass of Mg = 24 g/mol
Mole of Mg =?
Mole = mass /Molar mass
Mole of Mg = 37.2 / 24
Mole of Mg = 1.55 moles
Next, we shall write the balanced equation for the reaction. This is illustrated below:
3Mg + N₂ —> Mg₃N₂
From the balanced equation above,
3 moles of Mg reacted with 1 mole of N₂.
Therefore, 1.55 moles of Mg will react with = (1.55 × 1)/3 = 0.517 mole of N₂
Thus, 0.517 mole of N₂ is need for the reaction.
Finally, we shall determine the volume of N₂ needed for the reaction as follow:
Recall:
1 mole of a gas occupies 22.4 L at STP.
1 mole of N₂ occupied 22.4 L at STP.
Therefore, 0.517 mole of N₂ will occupy = 0.517 × 22.4 = 11.58 L at STP
Thus, 11.58 L of N₂ is needed for the reaction.
Answer:
0.5875L
Explanation:
concentration = mole/ volume
n(LiCl) = 20 / (7 + 35.5) = 0.47 mol
volume = mole / conc.
volume = 0.47 /0.8
= 0.5875 dm³ = 0.5875L
True
After the process of Ore processing, Enrichment, Fuel production and being passed through the reactor core the last remaining step is spent fuel disposal.
This is what i have found i hope this helps
